# $A,B,w \in \frac{\mathbb{Q}[t]}{(t^2-1)}[x,y]$: $\operatorname{Jac}(A,B)=1$, $\operatorname{Jac}(A,w)=0$, $w \notin \frac{\mathbb{Q}[t]}{(t^2-1)}[A]$

Let $$R=\frac{\mathbb{Q}[t]}{(t^2-1)}$$; trivially, $$R$$ is not an integral domain, since $$(\overline{t-1})(\overline{t+1})=\overline{t^2-1}=\overline{0}$$.

Is it possible to find $$A,B,w \in R[x,y]$$ such that the following three conditions are satisfied:

(i) $$\operatorname{Jac}(A,B)\in \{1,-1\}$$. (ii) $$\operatorname{Jac}(A,w)=0$$. (iii) $$w \notin R[A]$$.

Remarks:

• Similar questions are: 1, 2, 3, 4; especially question 3.

• Here we are not able to talk about normal or non-normal domains (in the simplest meaning), because now $$R$$ is not an integral domain.

• Notice that $$A=\overline{t−1}x+\overline{t}y$$, $$B=\overline{t}x+\overline{t+1}y$$, $$w=\overline{t+1}y$$ do not satisfy all the three conditions; indeed, conditions (i) and (ii) are satisfied, but condition (iii) is not, since $$R[A] \ni \overline{t+1}A=\overline{t+1}(\overline{t−1}x+\overline{t}y)= \overline{t^2-1}x+\overline{(t+1)t}y=\overline{0}x+\overline{t^2+t}y=\overline{1+t}y=w$$.

Thank you very much!